Circle Ques 13
- Let $O(0,0)$ and $A(0,1)$ be two fixed points, then the locus of a point $P$ such that the perimeter of $\triangle A O P$ is 4 , is
(2019 Main, 8 April I)
(a) $8 x^{2}-9 y^{2}+9 y=18$
(b) $9 x^{2}-8 y^{2}+8 y=16$
(c) $9 x^{2}+8 y^{2}-8 y=16$
(d) $8 x^{2}+9 y^{2}-9 y=18$
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Answer:
Correct Answer: 13.(c)
Solution:
Formula:
- Given vertices of $\triangle A O P$ are $O(0,0)$ and $A(0,1)$
Let the coordinates of point $P$ are $(x, y)$.
Clearly, perimeter $=O A+A P+O P=4$ (given)
$\Rightarrow \sqrt{(0-0)^{2}+(0-1)^{2}}+\sqrt{(0-x)^{2}+(1-y)^{2}}+\sqrt{x^{2}+y^{2}}=4$
$\Rightarrow 1+\sqrt{x^{2}+(y-1)^{2}}+\sqrt{x^{2}+y^{2}}=4$
$\Rightarrow \sqrt{x^{2}+y^{2}-2 y+1}+\sqrt{x^{2}+y^{2}}=3$
$\Rightarrow \sqrt{x^{2}+y^{2}-2 y+1}=3-\sqrt{x^{2}+y^{2}}$
$\Rightarrow \quad x^{2}+y^{2}-2 y+1=9+x^{2}+y^{2}-6 \sqrt{x^{2}+y^{2}}$
$ \quad $[squaring both sides]
$\Rightarrow \quad 1-2 y=9-6 \sqrt{x^{2}+y^{2}}$
$\Rightarrow \quad 6 \sqrt{x^{2}+y^{2}}=2 y+8$
$\Rightarrow \quad 3 \sqrt{x^{2}+y^{2}}=y+4$
$\Rightarrow \quad 9\left(x^{2}+y^{2}\right)=(y+4)^{2} \quad $ [squaring both sides]
$\Rightarrow \quad 9 x^{2}+9 y^{2}=y^{2}+8 y+16$
$\Rightarrow \quad 9 x^{2}+8 y^{2}-8 y=16$
Thus, the locus of point $P(x, y)$ is
$ 9 x^{2}+8 y^{2}-8 y=16 $
BETA
